Question:

At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to:

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When you know the amounts at two different times under simple interest, the difference of amounts directly gives you the interest for the extra period. From that, you can easily find the yearly interest, then the rate and principal, and finally plug those into a compound interest calculation.
Updated On: Dec 5, 2025
  • \(3096\)
  • \(3221\)
  • \(3180\)
  • \(3150\)
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The Correct Option is B

Solution and Explanation

Step 1: Find the principal and simple interest rate. Amount after 3 years: \[ A_1 = 13920. \] Amount after 6.5 years: \[ A_2 = 18960. \] Time difference: \[ 6.5 - 3 = 3.5 \text{ years}. \] Extra interest in these 3.5 years: \[ \Delta I = A_2 - A_1 = 18960 - 13920 = 5040. \] So simple interest per year: \[ \text{SI per year} = \frac{5040}{3.5} = 1440. \] Interest for first 3 years: \[ \text{SI}_{3\text{ yrs}} = 1440 \times 3 = 4320. \] Hence principal: \[ P = A_1 - \text{SI}_{3\text{ yrs}} = 13920 - 4320 = 9600. \] Rate of interest: \[ R = \frac{\text{SI per year}}{P} \times 100 = \frac{1440}{9600} \times 100 = 15% \text{ p.a.} \]
Step 2: Compound interest for 2 years, half-yearly. Principal: \[ P = 9600. \] Rate per half-year: \[ r = \frac{15%}{2} = 7.5% = 0.075. \] Number of half-yearly periods in 2 years: \[ n = 2 \times 2 = 4. \] Amount: \[ A = P\left(1 + \frac{r}{100}\right)^n = 9600 \times (1.075)^4. \] Approximate: \[ (1.075)^2 \approx 1.1556,\quad (1.1556)^2 \approx 1.335\ (\text{approximately}), \] so \[ A \approx 9600 \times 1.335 \approx 12820. \] Compound interest: \[ \text{CI} = A - P \approx 12820 - 9600 = 3220 \text{ (approximately)}. \] Among the given options, the nearest value is \(3221\). Therefore, the total interest would have been approximately: \[ \boxed{3221}. \]
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